$\sum\limits_{k=0}^{{29}}{{{-3(0.9)^{k}}}} \approx$ Choose 1 answer: Choose 1 answer: (Choice A) A $-4.63\cdot10^{13}$ (Choice B) B $ -28.7 $ (Choice C) C $ -1.65 $ (Choice D) D $ -0.14 $
Solution: What is the question asking for? The question is asking for the sum of the values of $-3(0.9)^k$ from $k = 0$ to $k = 29$ : $-3(0.9)^0 -3(0.9)^1 +... -3(0.9)^{29} $ The series is geometric because the formula $-3(0.9)^k$ is an exponential function of $k$. Formula for geometric series The sum $S_n$ of a finite geometric series is $S_n = \dfrac{a_1(1-r^n)}{1-r}$ where $a_1$ is the first term, $r$ is the common ratio, and $n$ is the number of terms. What do we need to use the formula? The number of terms $n$ is ${30}$ because there are ${30}$ numbers from $0$ to $29$. The first term $a_1$ is ${-3}$ because $-3(0.9)^0 = {-3}$. The common ratio $r$ is ${0.9}$ because it is the base of the exponent in $-3({0.9})^k$. Find the sum $(S_n)$ of the series $\begin{aligned} S_n &= \dfrac{a_1(1-r^n)}{1-r} \\\\ S_{{30}}&=\dfrac{{-3}(1-\left({0.9}\right)^{{30}})}{1-\left({0.9}\right)} \\\\ S_{{30}}&=-30{(1-\left({0.9}\right)^{{30}})} \\\\ S_{{{30}}} &\approx -28.73 \end{aligned}$ The answer $ -28.7 $